Recent MathOverflow Questions

Representability of integral Kähler forms

Math Overflow Recent Questions - Sat, 07/21/2018 - 04:24

Let $X$ be a complex manifold admitting Kähler forms. If $X$ is compact and $\omega \in H^2(X,\mathbb{Z})$ is an integral Kähler two-form, there exists a Hermitian holomorphic line bundle $(L,H)$ such that:

$\omega = \frac{i}{2\pi} F_{H}$

where $F_{H}$ is the curvature of the Chern connection associated to the Hermitian structure $H$.

However, if $X$ is not compact and $\omega \in H^2(X,\mathbb{Z})$ is an integral Kähler two-form, then, and as far as I know, the above may not be true and such $(L,H)$ may not exist. I fail to find examples that realize this possibility, for $X$ non-compact, so any help is welcome.


Combinatorics of $p$-Kazhdan--lusztig polynomials

Math Overflow Recent Questions - Sat, 07/21/2018 - 03:13

When can we (and can we not!) understand the dimensions of simple modules, $D(\lambda)$, of symmetric groups in a combinatorial fashion?

Let's assume that I'm going to try to do this using the theory of $p$-Kazhdan-Lusztig polynomials and that I want to do this in the language of tilting modules.

Then my question becomes, given $\lambda$ a partition of $n$ with at most $h$ columns: when can I combinatorially understand both $T(\lambda)$ and $T(\mu)$ for all partitions $\mu$ of $n$ which are less dominant than $\lambda$ (in the dominance ordering on the block).

I believe it's true to say that we cannot hope to understand the $p$-canonical basis in a combinatorial fashion in full generality. I believe that would contain (as a subproblem) certain number theoretic questions which are not expected to have combinatorial solutions. But this was just word of mouth, does anyone have a reference for this?

On the other hand, Lusztig--Williamson have a conjecture for tilting modules from the first $p^3$-alcove for $\rm GL_3$ in terms of the combinatorics of ``billiards". If true, this means we can understand ${\rm Dim}_k(D(\lambda))$ providing $\lambda$ has 3 columns and is a partition of weight at most $p^2$(ish).

Now for the general case. It seems fair (given the above conjecture of Lusztig--Williamson and assuming the word-of-mouth citation above is correct) to say that at the moment, the only general class of $p$-Kazhdan--Lusztig polynomials we can hope to combinatorially understand is those which coincide with the actual good old fashioned Kazhdan--Lusztig polynomials.

So.... when does this happen? Is it correct to say that a necessary condition for this is that we are in the first $p^2$-alcove (weight of $\lambda$ is less than $p$)? Now, my reading of Riche--Williamson's result is that each alcove has its own bound on the prime. But then, is there a sensible way of talking about my question -- which involved not only understanding a given partition $\lambda$ in an alcove $A_\lambda$ but all partitions $\mu \vdash n$ such that $\mu<\lambda$ in alcoves $A_\mu$? I'm not necessarily looking for a technical condition, but more of a feeling for ``how deep into the first $p^2$-alcove can we go?".

In particular, we can't go as far as the whole of the first $p^2$-alcove because that would prove Andersen's conjecture (which Riche--Williamson explicitly state that they do not do). But can one think somehow that as $p\rightarrow \infty$ that we get most of the the first $p^2$-alcove in some sense and that it's just a few partitions ``near" the $p^2$-wall for which the result fails?

How can least squares regression be modified to penalize errors more heavily for small values?

Math Overflow Recent Questions - Sat, 07/21/2018 - 03:09

As usual, $f(x_i)$ is some linear combination of the variables, with the error total:

$\sum_{i=1}^N (y_i - f(x_i))^2$

I would like to penalize errors near 0 more heavily. One conceivable approach is to use a sigmoid function e.g. $sig(x) = tanh(\alpha x)$ so the error total becomes:

$\sum_{i=1}^N (sig(y_i) - sig(f(x_i)))^2$

The parameter $\alpha$ gives a nice amount of control - the lower the value, the less effect the sigmoid has.

So I really don't know how to proceed with this, and if minimizing this error can be done as quickly and accurately as vanilla Least Squares. Maybe there are other approaches? Thanks for any help - I've never done well with stats!

Generalizations of Pedal Coordinates

Math Overflow Recent Questions - Sat, 07/21/2018 - 02:53

I recently "stumbled upon" the article Pedal coordinates, Dark Kepler and other force problems by Petr Blaschke from 2017; further search about Pedal Coordinates didn't bring up any other relevant online resources; from the references on Wolfram Mathworld I get the impression, that the topic hasn't been the subject and/or means of mathematical research for some decades.


  • where, except in the cited article by Petr Blaschke, have Pedal Coordinates played an important role in recent mathematical publications

  • have there been attempts to generalize Pedal Coordinates to higher dimensions and/or other numbers besides the reals like e.g. complex numbers

Find a unique oredered pair for n<128 [on hold]

Math Overflow Recent Questions - Sat, 07/21/2018 - 02:18

Suppose that $n<128$ is a natural number. Do we find a unique ordered pair $(i,j)$ such that $i,j$ are natural numbers less than 7, for n? Or do we find a unique ordered $(i,j,k)$, such that $i,j,k<7$, for n? We know that with Chinese reminder theorem, we can find a unique ordered pair $(i,j)$ for n. But I mean, without using this theorem, do we find such a unique ordered pair for any natural number less than 128?

In other words, for natural number $n<128$, there exists a unique $(i,j)$ such that $i,j<7$ are natural numbers and vice versa? Or do there exist a unique $(i,j,k)$, such that $i,j,k<7$ for $n<128$?

Slove it with your Mind - #Paul Ardaji [on hold]

Math Overflow Recent Questions - Sat, 07/21/2018 - 01:57

Given a triangle ABC. BL is the bisector of angle ABC, H is the orthocenter and P is the mid-point of AC. PH intersects BL at Q. If ∠ABC=β, find the ratio PQ:HQ.If QR⊥BC and QS⊥AB, prove that the orthocenter lies on RS.

  • By Paul Ardaji

A balanced tree-like presentation of $S_3$

Math Overflow Recent Questions - Sat, 07/21/2018 - 01:54

Does the 6-element group $S_3$ have a finite (balanced) semigroup presentation of the form $$\langle a_1,...,a_n\mid a_1=u_1, a_2=u_2,...,a_n=u_n\rangle$$ where $u_1,u_2,...,u_n$ are semigroup words? Let us call such a semigroup presentation {\it tree-like}.

Edit: By a semigroup word, I mean a word without inverses of letters. By a semigroup presentation, I mean a presentation in the class of semigroups (not groups). For example, $\langle a,b\mid a^3=a, b=ab^2a\rangle$ defines $S_3$ in the class of groups but a "bigger" semigroup in the class of semigroups. In that semigroup, $b$ does not divide $a$, so it is not a group.

Motivation 1. The cyclic group of order $n$ has a tree-like semigroup presentation $\langle a\mid a=a^{n+1}\rangle$. Matt Brin noted that the 8-element quaternion group has the tree-like presentation $\langle a,b,c \mid a=bc, b=ca, c=ab\rangle$. The groups $S_m, m>3$ do not have tree-like presentations because they do not even have balanced (same number of relations and generators) presentations at all since their Schur multipliers are non-trivial. The group $S_3$ has a balanced presentation.

Motivation 2. Every tree-like semigroup presentation corresponds to a (closed) subgroup of the R. Thompson group $F$. For example as shown by Guba the Brin tree-like presentation of the quaternion group corresponds to a copy of the Thompson group $F_9$ (the group of piecewise linear homeomorphisms of $[0,1]$ with slopes of the form $9^k$ and break points of the derivative from $\mathbb{Z}[1/9]$).

How to visualize the Frobenius endomorphism?

Math Overflow Recent Questions - Sat, 07/21/2018 - 01:11

As the question title asks for, how do others "visualize" the Frobenius endomorphism? I asked my advisor and he said he didn't know and that I could go and ask on MO and possibly get miseducated. So that's what I am doing. Bonus points for pictures.

Reference request: basics about modular curves

Math Overflow Recent Questions - Sat, 07/21/2018 - 01:07

Where can I find a reference (with carefully written proofs) for basic facts about modular curves? Namely:

  • Congruence subgroups

  • The open modular curve $Y_\Gamma$ admits the structure of a Riemann surface (dealing with the issue of orbifold points)

  • The curve $Y_\Gamma$ admits a completion at the cusps to give a closed Riemann surface $X_\Gamma$

Ideally, with no mention of the word "stack". I'm mostly interested in understanding the complex analytic side of the story.

Statistical independence of eigenvectors of real symmetric Gaussian random matrices

Math Overflow Recent Questions - Fri, 07/20/2018 - 23:13

What is known about the statistical independence of the eigenvectors of a real symmetric matrix with independent Gaussian entries with zero mean, and finite variance? The matrix elements are not assumed to have same variance.

I see some results for Wigner matrices in literature, where the entries are i.i.d. standard Gaussian (except diagonal) - though even in this case, whether the eigenvectors are in fact statistically independent is not entirely clear to me (though I suspect that to be the case for Wigner matrices).

So, does there exist results regarding statistical independence of eigenvectors of random real symmetric matrices with non-identical, but statistically independent Gaussian entries? Any references for this in literature would be helpful.

A generalization of Chebyshev's sum inequality

Math Overflow Recent Questions - Fri, 07/20/2018 - 22:45

From some my previous questions here and here and well-known rearrangement inequality. I pose an inequality as follows and I am looking for the proof or a reference.

Inequality: Let $y=f(x,y)$ is positive continuous function in interval $[a, b] \times [c, d]$. Such that: $f'_x(x,y)>0$ and $f'_y(x,y)>0$ in $[a, b] \times [c, d]$. If $a \le x_1 \le x_2 \le \cdots \le x_n \le b$ and $c \le y_1 \le y_2 \le \cdots \le y_n \le d$ then

$$n\sum_{i=1}^{n}f(x_i, y_i) \ge \sum_{i=1}^{n}\sum_{j=1}^{n}f(x_i, y_j)$$

if If $a \le x_1 \le x_2 \le \cdots \le x_n \le b$ and $d \ge y_1 \ge y_2 \ge \cdots \ge y_n \ge c$ then

$$n\sum_{i=1}^{n}f(x_i, y_i) \le \sum_{i=1}^{n}\sum_{j=1}^{n}f(x_i, y_j)$$

EDIT: Allow me adding condition $f''_{xy}>0$ $f''_{yx}>0$

Remarks: The result be can general to $n$ variant.

Any 3-manifold can be realized as the boundary of a 4-manifold

Math Overflow Recent Questions - Fri, 07/20/2018 - 20:48

We know

"Any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$." See this post: Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold?

I heard this statement is true:

  • (1) Any closed 3-manifold is a boundary of some compact 4-manifold.

See also this paper p.2's 3rd paragraph uses the fact:

  • (2) Any 3-manifold $M$ can be realized as the boundary of a 4-manifold $B$.

In particular, we know that all 3-manifolds can be triangulable. However for 4-manifolds, there are simply connected non-triangulable manifolds (such as the E$_8$ manifold). (Note: a closed 4-manifold is triangulable if and only if it's smoothable.) See this MO post: Not all manifolds can be triangulated

  • (3) For any 3-manifold $M_3$ that can be realized as the boundary of a 4-manifold $B_4$, the $M_3$ must be triangulable. So must the $M_3$ be the boundary of a triangulable 4-manifold $B_4$?

  • (4) Are there any non-triangulable 4-manifold $B_4'$ with a 3-dimensional boundary (i.e. $B_4'$ is not closed)? Then would the 3-manifold boundary $M_3'$ be triangulable (if $M_3'$ is non-triangulable, isn't that leads to a contradiction)?

Can one show these (1), (2) and explain them as intuitively as possible?

Is the exponential version of Catalan-Dickson conjecture true?

Math Overflow Recent Questions - Fri, 07/20/2018 - 15:44

The aliquot sum function $s:\mathbb{N}\rightarrow \mathbb{N}$ assigns to any natural number $n$ the sum of its proper divisors. Perfect numbers are fixed points of this function. The open conjecture of Catalan and Dickson predicts the behavior of $\{s^{(k)}(n)\}_{k\in \omega}$, the aliquot sequence of $n$ produced by the infinite iteration of $s$ on $n$. The conjecture states that for any natural number $n$, $\{s^{(k)}(n)\}_{k\in \omega}$ ends up becoming constant on $0$ or a perfect number or circulating in a loop of sociable numbers.

Now, let's assume an analogy of this conjecture for the exponentiation operator rather than the multiplication.

Definition. For natural numbers $n, c, d$:

(1) $c$ and $d$ are exp-divisors of $n$ if $c^d=n$. Any of them is called proper if it is not equal to $n$.

(2) The aliquot product function $t:\mathbb{N}^{>1}\rightarrow \mathbb{N}$ assigns to any natural number $n>1$ the product of its proper exp-divisors.

(3) The exp-aliquot sequence of $n$ is $\{t^{(k)}(n)\}_{k\in \omega}$.

(4) An exp-perfect number is a fixed point of $t$.

(5) A finite sequence of natural numbers is called exp-sociable if they form an aliquot product loop. (i.e. Any of them is the aliquot product of the previous one and the first number is the aliquot product of the last one).

Example. We have $3^2=9^1$ so $1, 2, 3$ are proper exp-divisors of $9$ and so $t(9)=2.3=6$. As $6$ has no proper exp-divisors except $1$ so $t^2(9)=t(6)=1$.

Remark. Note that just like the case of abundant numbers, there are natural numbers whose aliquot product is greater than the number itself. For instance, we have $2^8=4^4=16^2=256^1$. Thus the proper exp-divisors of $256$ are $1, 2, 4, 8, 16$ and so $t(256)=1024$. But even in this case the aliquot sequence eventually terminates at $1$ because we have $t^3(256)=t^2(1024)=t(12800)=1$. This observation gives rise to the following questions:

Question 1. Does $t$ have any fixed points? In other words, is there any exp-perfect number?

Question 2. Given a natural number $k\geq 2$, is there a sequence of exp-sociable numbers of length $k$?

Question 3. Is the exponential version of Catalan-Dickson conjecture true? Precisely, is it true that for any natural number $n>1$, the sequence $\{t^{(k)}(n)\}_{k\in \omega}$ ends up becoming constant on $1$ or an exp-perfect number or falling into a loop of exp-sociable numbers?

Note that assuming a negative answer to the questions 1 and 2 (which sounds likely to me), the question 3 will have the following neat formulation: $\forall n>1~~\exists k\geq 1~~~t^{(k)}(n)=1$ which shows a greater degree of nice behavior in comparison with the case of classical aliquot sequences.

Optimization problem on trace with both the positive semi definite and non positive semidefinite matrix

Math Overflow Recent Questions - Fri, 07/20/2018 - 09:30

Given two $N \times N$ symmetric matrices $A, B$, where $A$ is positive semidefinite while $B$ is not positive semidefinite. I am interested in solving unitary constrained trace maximization problem:

$\arg \max_{U}\text{trace}(UAU^TB),\;\;$ subject to $\;U^TU=I.$

When the matrix $B$ is positive semidefinite and diagonal the solution set $U$ will be the ordered eigenvectors of $A$. But in my case the matrix $B$ is surely nonpositive definite.

Existence of an antiderivative function on an arbitrary subset of $\mathbb{R}$

Math Overflow Recent Questions - Fri, 07/20/2018 - 09:27

Let $f:\mathbb{R}\to \mathbb{R}$ be continuous at $x$ for every $x\in I$ where $I\subset \mathbb R$ could be arbitrary. Does there always exist a function $F:\mathbb{R}\to \mathbb{R}$ differentiable on $I$ and $F'(x) = f(x)$ for every $x \in I$?

The definition of a primitive is naturally defined on an interval. what sort of weaker result can we obtain under weaker hypotheses?.

  • If I is an interval or an open set, the answer to the question is positive.

  • If f is locally Lebesgue integrable,the answer to the question is
    also positive.

I have already asked the question here

Parabolic variational inequality: regularity of the time derivative in $L^2(0,T;H)$?

Math Overflow Recent Questions - Fri, 07/20/2018 - 08:30

Let $V \subset H \subset V^*$ be a Gelfand triple of Hilbert spaces. Take $f,\psi \in L^2(0,T;H)$ and consider the VI: find $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V^*)$ such that

$$u(t) \leq \psi(t) : \int_0^T \langle u' + Au - f, v-u \rangle \geq 0\quad \forall v \in L^2(0,T;V) : v(t) \leq \psi(t).$$

$A$ is some smooth elliptic operator.

Can I expect $u' \in L^2(0,T;H)$, at least with additional assumptions? Could anyone refer me to a reference for this? The only literature is see has the obstacle $\psi$ independent of time, which is not the case here.

Create approximations of finite integer sequence

Math Overflow Recent Questions - Fri, 07/20/2018 - 07:59

I am looking for a representation of approximations for a finite but very long integer sequence. The actual sequence has on the order of $2^{150}$ elements, so it would be impossible to calculate or store the entire sequence. The plan is to create a program that can come up with approximations of the sequence and through evolution slowly improve the approximation as it is running.

What I have:

  • I would start with small and random approximations.
  • I have a good test for comparing two approximations to see which one is better. Better approximations will "mate" with each other and generate new approximations.
  • In addition, some random changes will be introduced as "mutations" for some of the children.

Things I have considered:

  • Polynomials, but they are not going to give me integers, and also have the downside that they will be computationally expensive to when they reach high orders, which they quickly would.

  • I looked into a Mersenne Twister, but this will probably never yield good approximations.

  • I have also considered a net of NAND gates. I would start with a small random net and let the evolution reuse portions of the net in subsequent generations. This is the best I can come up with on my own.

What type of structure or formula can approximate a finite integer sequence to arbitrary precision, while at the same time be fast to evaluate? The approximations does not need to be finite, but they must start at zero and be defined and computable for every integer input up to $2^{150}$ or so.

A rearrangement inequality for exponentiation function

Math Overflow Recent Questions - Fri, 07/20/2018 - 03:56

Inequality: If $n$ be positive integer $n \ge 2$ and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n \ge 0$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \ge \Pi_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

If $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \cdots \le \alpha_n$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \le \Pi_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

Equality holdes if only if $a_1=a_2=a_3=...=a_n$

By my computation the inequality as follows good be true for case $n=2$ and $n=3$. So I conjecture the inequality above true with $n \ge 2$.

Case n=2:

If $a_1 \ge a_2 \ge 0$ and $\alpha_1 \ge \alpha_2 \ge 0$ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \ge (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$


If $a_1 \ge a_2 \ge 0$ and $0 \le \alpha_1 \le \alpha_2 $ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \le (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

Case n=3:

If $a_1 \ge a_2 \ge a_3 \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \alpha_3 \ge 0$ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2}+a_3^{\alpha_3})^3 \ge (a_1^{\alpha_1}+a_1^{\alpha_2}+a_1^{\alpha_3})(a_2^{\alpha_1}+a_2^{\alpha_2}+a_2^{\alpha_3})(a_3^{\alpha_1}+a_3^{\alpha_2}+a_3^{\alpha_3})$$


If $a_1 \ge a_2 \ge a_3 \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \alpha_3 $ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2}+a_3^{\alpha_3})^3 \le (a_1^{\alpha_1}+a_1^{\alpha_2}+a_1^{\alpha_3})(a_2^{\alpha_1}+a_2^{\alpha_2}+a_2^{\alpha_3})(a_3^{\alpha_1}+a_3^{\alpha_2}+a_3^{\alpha_3})$$

My question: Could You give a comment, a reference or a proof?

Rotation invariant of surface

Math Overflow Recent Questions - Fri, 07/20/2018 - 02:11

Let $(x, y, f(x,y))$ be a surface in $\mathbb{R^3}$. It is written in a book without proof that all rotation invariant (rotating around $z$-axis) of $f$ are combinations of the following four quantities:

  • $\Delta f = f_{xx}+f_{yy}$
  • $||\nabla f||^2=f_x^2 + f_y ^2$
  • $\det{H} = f_{xx} f_{yy} - f_{xy}^2$
  • $(\nabla f, H \nabla f) = f_{xx}f_x^2 + 2f_{xy}f_xf_y+f_{yy}f_y^2$

How do I prove this?

Definition of rotation invariant

My book doesn't provide a precise definition of rotation invariant but I think this is defined as follows:

Let $R[F]={C^{\infty}(\mathbb{R}^2)}[F_{x}, F_{y}, F_{xx}, F_{xy}, F_{yy}, ...]$ be a polynomial ring of infinite variables and $R(F)$ be its field of fractions. Elements in $R(F)$ can be seen as functionals. For a element $H(F)\in R(F)$, we define that $H(F)$ is rotation invariant iff for any rotation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$around the origin and for any $f\in C^{\infty}(\mathbb{R}^2)$, $H(f\circ T)=H(f)\circ T$ holds.

Definition of combinations

Let $S(F)$ be a field of fractions of the ring $C^{\infty}(\mathbb{R}^2)[\Delta F, ||\nabla F||, \det{H},(\nabla F, H \nabla F)]$. A combination of the four quantities means an element of $S(F)$.

Conic bundles which are rational but with non-rational generic fibres

Math Overflow Recent Questions - Thu, 07/19/2018 - 15:03

Let $n\ge 1$ be an integer and let us work over the field of complex numbers. Let $\mathcal{R}_n$ denote the set of rational conic bundles $\pi\colon X\to \mathbb{P}^n$ (morphisms such that the generic fibre is a geometry irreducible conic and $X$ is rational), up to square equivalence: two conic bundles $\pi_1\colon X_1\to \mathbb{P}^n$ and $\pi_2\colon X_2\to \mathbb{P}^n$ are equivalent if and only if there are birational maps $\psi\colon X_1 \dashrightarrow X_2$ and $\varphi\colon \mathbb{P}^n\dashrightarrow \mathbb{P}^n$ such that $\varphi\circ \pi_1=\pi_2\circ \psi$. This corresponds to have a commutative diagram (where horizontal arrows are birational) $\require{AMScd}$ \begin{CD} X_1 @>\psi>> X_2\\ @V \pi_1 V V @VV \pi_2 V\\ \mathbb{P}^n @>>\varphi> \mathbb{P}^n. \end{CD}

If $n=1$, then $\mathcal{R}_n$ consists of a single point (result of Enriques).

Question: Is the set $\mathcal{R}_n$ uncountable for $n\ge 2$?

For $n=2$, the answer is yes, as we can take the discriminant locus to be any smooth irreducible cubic, and these are pairwise not equivalent under birational maps of $\mathbb{P}^2$ if they are not isomorphic.

For $n>2$, I guess that the answer is again yes, but I do not know any reference for this.


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